These formulas are easy to memorize using a tool called the “del”
operator, denoted by the nabla symbol $\nabla$. Think of $\nabla$ as a “fake” vector
composed of all the partial derivatives that we use just to help us
remember the formulas:
\begin{align*}
\nabla = \left(\pdiff{}{x}, \pdiff{}{y}, \pdiff{}{z}\right).
\end{align*}
Although it may not seem to make sense to just have the partial
derivatives without them acting on a function, we won't worry about
that. This is just notation.

Now, let's take the dot product of the $\nabla$ vector with
$\dlvf=(\dlvfc_1,\dlvfc_2, \dlvfc_3)$:
\begin{align*}
\nabla \cdot \dlvf &= \left(\pdiff{}{x}, \pdiff{}{y},
\pdiff{}{z}\right)
\cdot (\dlvfc_1,\dlvfc_2, \dlvfc_3)\\
&= \pdiff{}{x}\dlvfc_1 + \pdiff{}{y}\dlvfc_2 +
\pdiff{}{z}\dlvfc_3
\end{align*}
If we think of each “multiplication” in the dot product as instead
being the derivative of the corresponding $\dlvfc$, then we have the
formula for the divergence. So, if you can remember the del operator
$\nabla$ and how to take a dot product, you can easily remember the
formula for the divergence
\begin{align*}
\div \dlvf = \nabla \cdot \dlvf =
\pdiff{\dlvfc_1}{x} + \pdiff{\dlvfc_2}{y} + \pdiff{\dlvfc_3}{z}.
\end{align*}

This notation is also helpful because you will always know that
$\nabla \cdot \dlvf$ is a scalar (since, of course, you know that the
dot product is a scalar product).

This is exactly the formula we gave above. So if you can use the rule
that “multiplication” by $\pdiff{}{x}$ is the same as taking the
partial derivative with respect to $x$ (and similar for the other
derivatives), then you can remember the curl formula by
\begin{align*}
\curl \dlvf = \nabla \times \dlvf.
\end{align*}